Characteristics of Feynman diagrams for correlators at a given order of lambda

[Mithra]

I've been given an action of

$S = -\frac{1}{2}m^2\phi^2 - \frac{\lambda}{3!}\phi^3$

and the question then require drawing the feynman diagrams for the $<\phi\phi>$ at order $\lambda^2$.
I am told that there are 2 connected diagrams at one-loop order and one disconnected two-loop diagram and asked to draw them, however I am unsure what exactly defines which order of lambda a feynman diagram will be? As there is only a phi^3 in the action I believe this means that there can only be three-point interaction vertexes rather than four but I'm not sure if that's correct?

I think my connected diagrams are ok however for the two-loop diagram I initially had a single line connecting two points and then separately a point with two loops coming from it, however looking in the notes I see that that is a first order diagram (and also, possibly requires a phi^4 as there are four connections to the single point from the two loops?).

If anyone has any pointers as to what I should be looking for, or if I'm totally on the wrong track, that would be brilliant thanks!

 

[kevinferreira]

The order of lambda gives you the number of vertex (interaction points) of the diagrams you have to draw. In this case, 2. The phi^3 interaction term gives you, as you said, the number of lines converging to a vertex (interacting fields).

Your second paragraph on the disconnected 2-loop diagram shows that you see that what you did is incorrect, you have to review the way you did it. Did you learn how to compute time ordered correlation functions in general? (Using Dyson's series and the interacting vacuum).
This gives you all diagrams directly, even if it is sometimes too fastidious to be done.

With respect to the disconnected diagram, it will have to include a line for the two fields, and a vacuum bubble (the name of the closed disconnected to the fields parts of the disconnected diagrams)... Does this name gives you any hint? You have to draw a closed diagram, with its vertex's connecting 3 lines, and 2 vertex's in total... Think about it.

 

[Mithra]

The order of lambda gives you the number of vertex (interaction points) of the diagrams you have to draw. In this case, 2. The phi^3 interaction term gives you, as you said, the number of lines converging to a vertex (interacting fields).

Your second paragraph on the disconnected 2-loop diagram shows that you see that what you did is incorrect, you have to review the way you did it. Did you learn how to compute time ordered correlation functions in general? (Using Dyson's series and the interacting vacuum).
This gives you all diagrams directly, even if it is sometimes too fastidious to be done.

With respect to the disconnected diagram, it will have to include a line for the two fields, and a vacuum bubble (the name of the closed disconnected to the fields parts of the disconnected diagrams)... Does this name gives you any hint? You have to draw a closed diagram, with its vertex's connecting 3 lines, and 2 vertex's in total... Think about it.

Awesome, thanks very much. Would a diagram with a single line for the two fields and a 'bubble' as a line with each end vertex having a loop be valid then?

For the next part a similar question is asked with a phi^3 correlator and two-loop disconnected diagrams at order $\lambda^3$. With the action as above does every vertex have to be three fields, or is it up to three fields? The only way I can see of drawing double loop diagrams involves two fields interacting with a loop in the middle or emitting a 'particle' which then becomes a loop, so both using up two of the three vertexes, after that I can only think of having a single 'vertex' which is a loop but that would only have two connections and also does that count as a 'field', if not there are only two fields in the diagram...

Thanks for the help!

 

[kevinferreira]

Awesome, thanks very much. Would a diagram with a single line for the two fields and a 'bubble' as a line with each end vertex having a loop be valid then?

For the next part a similar question is asked with a phi^3 correlator and two-loop disconnected diagrams at order $\lambda^3$. With the action as above does every vertex have to be three fields, or is it up to three fields? The only way I can see of drawing double loop diagrams involves two fields interacting with a loop in the middle or emitting a 'particle' which then becomes a loop, so both using up two of the three vertexes, after that I can only think of having a single 'vertex' which is a loop but that would only have two connections and also does that count as a 'field', if not there are only two fields in the diagram...

Thanks for the help!

Hum, yeah, I think it would be a valid diagram. I was thinking about another one, a circle with a diameter line, which is popular in QED (qed is a phi^3 theory, except that it deals with two fermionic electron-positron fields and a bosonic vectorial photon field). But anyway, there are a lot of disconnected diagrams like this, but we don't care about them when computing things, they vanish.

With the action as above, every vertex has 3 lines and only 3 lines. The quadratic terms on the Lagrangian give the propagators, the 'inertia' of the fields. It doesn't exist vertices with 2 fields, a field can't simply interact with itself and suddenly become another type of field.

I didn't get your last explanation. When you say "phi^3 correlator" you mean a 3-point correlation function $\langle\phi\phi\phi\rangle$ at third order?

 

[Mithra]

Ah yes, that was the other one I was thinking of, thanks :). Of course yeah thinking about it in terms of actual fields rather than just maths a two field vertex wouldn't make much sense would it! :P

And yeah that's what I was referring to sorry, always a mistake not to use latex!

 

[kevinferreira]

Ah yes, that was the other one I was thinking of, thanks :). Of course yeah thinking about it in terms of actual fields rather than just maths a two field vertex wouldn't make much sense would it! :P

And yeah that's what I was referring to sorry, always a mistake not to use latex!

Okay, so a 2-loop disconnected diagram with 3 external fields at order 3. Well, you can simply put the 3 fields interacting trivially through a vertex (2-to-1 or 1-to-2, doesn't matter) and then the same bubble as in the precedent case, the line with two loops at each end as you said or the circle with line diameter as I said, or any other of the kind. They have 2 loops as required and 2 vertices, which together with the first one gives you 3, as we want it, 3rd order, disconnected.

Can you think about what would happen with the 2nd order term of $\langle\phi\phi\phi\rangle$ ?